Let P be any point on the line x – y + 3 = 0 and A be a fixed point (3, 4). If the family of lines given by the equations $(3 \sec θ + 5 \cos e c θ) x + (7 \sec θ - 3 \cos e c θ) y + 11 (\sec θ - \cos e c θ) = 0$ are concurrent at a point B for all permissible value of $θ,$ then which of the following option(s) is (are) TRUE?

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Maximum value of $| P A - P B |$ is $2 \sqrt{10} .$

Product of the abscissa and ordinate of point B is equal to 2

Minimum value of PA + PB is $2 \sqrt{34} .$

Sum of the abscissa and ordinate of point B is equal to – 1

answer is A, C.

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Detailed Solution

$(3 \sec θ + 5 \cos e c θ) x + (7 \sec θ - 3 \cos e c θ) y + 11 (\sec θ - \cos e c θ) = 0$
$\sec θ (3 x + 7 y + 11) + \cos e c θ (5 x - 3 y - 11) = 0$
Hence, family of lines are concurrent at the point of intersection of
3x + 7y + 11 = 0 and 5x – 3y – 11 = 0
Hence point B is (1, – 2).